\mathbb{Q}), E[p])$ be nontrivial?

9 events

when toggle format	what		by	license	comment
Nov 12, 2014 at 14:37	comment	added	Chris Wuthrich		Yes that is possible for $p=5$. It turns out to be equivalent to be a quadratic twist by 5 of a curve with a rational 5-torsion point.
Nov 12, 2014 at 8:56	comment	added	Ahmed Matar		@TylerLawson thanks for this very interesting example!
Nov 12, 2014 at 8:55	vote	accept	Ahmed Matar
Nov 11, 2014 at 23:33	comment	added	Tyler Lawson		@ChrisWuthrich Even worse, if I'm calculating correctly (based on your argument) there only appears to be a candidate subgroup of $GL_2(\Bbb F_p)$ which can support a nonzero cohomology group if $p \not \equiv 1 \mod 3$; the subgroup needs to be the set of matrices of the form $(\begin{smallmatrix}c^2 & * \\ 0 & c\end{smallmatrix})$. I don't know whether $p=5$ supports a curve with this type of torsion.
Nov 11, 2014 at 22:20	comment	added	Chris Wuthrich		You are right. More generally, If $G$ is the group of all matrices of the form $(\begin{smallmatrix} 1 & * \\ 0 & * \end{smallmatrix})$, then $H^1(G,E[p])= \mathbb{F}_p$ if $p =3$ and it is zero if $p>3$. Your example, and many other curves with a 3-torsion point rational over $\mathbb{Q}$, has indeed this group.
Nov 11, 2014 at 17:12	history	undeleted	Tyler Lawson
Nov 11, 2014 at 17:12	history	edited	Tyler Lawson	CC BY-SA 3.0	added 586 characters in body
Nov 11, 2014 at 16:37	history	deleted	Tyler Lawson		via Vote
Nov 11, 2014 at 16:17	history	answered	Tyler Lawson	CC BY-SA 3.0